Let $Z_{2^\infty}:=\{z\in \mathbb C:z^{2^n}=1,$for some $n\in \mathbb N\}$. This is a countable group. But I am not sure about the cardinality of the set of its subgroups. Does it have uncountably many subgroups. If yes, then what is the collection and how to find it?
I am also looking for an easy example of a countable group with uncountably many distinct subgroups. In stack, I have found groups that are a little non-obvious. Can someone help me a bit on this?
The Prüfer $p$-group has exactly one subgroup of order $p^n$ for each $n$. Those are the only subgroups. So it has countably many proper subgroups, plus the whole group. This gives countably many subgroups. The subgroups are exactly the set of elements satisfying $z^{p^n}=1$ for a fixed $n$.
For a countable group with uncountably many subgroups, consider the group $\oplus_{n=1}^{\infty}C_2$; then each subset $S$ of $\mathbb{N}$ determines a subgroup (the one in which the coordinates outside of $S$ are always trivial).
Another example is $\mathbb{Q}$ as an additive group. For every set $P$ of primes, let $H_P$ be the subgroup of rationals $q$ with the following property: if we write $q=\frac{a}{b}$ with $\gcd(a,b)=1$, and $p$ is a prime that divides $b$, then $p\in P$. Different sets of primes determine different subgroups, and there are uncountably many sets of primes.